我有兴趣查找自上次活动每个ID 以来的天数。数据如下所示：
df <- data.frame(date=as.Date(
c("06/07/2000","15/09/2000","15/10/2000","03/01/2001","17/03/2001",
"06/08/2010","15/09/2010","15/10/2010","03/01/2011","17/03/2011"), "%d/%m/%Y"),
event=c(0,0,1,0,1, 1,0,0,0,1),id = c(rep(1,5),rep(2,5)))
date event id
1 2000-07-06 0 1
2 2000-09-15 0 1
3 2000-10-15 1 1
4 2001-01-03 0 1
5 2001-03-17 1 1
6 2010-08-06 1 2
7 2010-09-15 0 2
8 2010-10-15 0 2
9 2011-01-03 0 2
10 2011-03-17 1 2
我从数据表解决方案here大量借用，但这不考虑ID。
library(data.table)
setDT(df)
setkey(df, date,id)
df = df[event == 1, .(lastevent = date), key = date][df, roll = TRUE]
df[, tae := difftime(lastevent, shift(lastevent, 1L, "lag"), unit = "days")]
df[event == 0, tae:= difftime(date, lastevent, unit = "days")]
它会生成以下输出
date lastevent event id tae
1: 2000-07-06 <NA> 0 1 NA days
2: 2000-09-15 <NA> 0 1 NA days
3: 2000-10-15 2000-10-15 1 1 NA days
4: 2001-01-03 2000-10-15 0 1 80 days
5: 2001-03-17 2001-03-17 1 1 153 days
6: 2010-08-06 2010-08-06 1 2 3429 days
7: 2010-09-15 2010-08-06 0 2 40 days
8: 2010-10-15 2010-08-06 0 2 70 days
9: 2011-01-03 2010-08-06 0 2 150 days
10: 2011-03-17 2011-03-17 1 2 223 days
我想要的输出如下：
date lastevent event id tae
1: 2000-07-06 <NA> 0 1 NA days
2: 2000-09-15 <NA> 0 1 NA days
3: 2000-10-15 2000-10-15 1 1 NA days
4: 2001-01-03 2000-10-15 0 1 80 days
5: 2001-03-17 2001-03-17 1 1 153 days
6: 2010-08-06 2010-08-06 1 2 NA days
7: 2010-09-15 2010-08-06 0 2 40 days
8: 2010-10-15 2010-08-06 0 2 70 days
9: 2011-01-03 2010-08-06 0 2 150 days
10: 2011-03-17 2011-03-17 1 2 223 days
唯一的区别是第6行的NA和 tae 列。 This是一个没有答复的相关帖子。我看过here，但解决方案在我的情况下不起作用。还有许多其他类似的问题但不适用于每个ID的计算。谢谢！
答案 0 :(得分：2)
df <- data.table(date=as.Date(c("06/07/2000","15/09/2000","15/10/2000","03/01/2001","17/03/2001","06/08/2010","15/09/2010","15/10/2010","03/01/2011","17/03/2011"),
"%d/%m/%Y"), event=c(0,0,1,0,1, 1,0,1,0,1),id = c(rep(1,5),rep(2,5)))
tempdt <- df[event==1,]
tempdt[,tae := date - shift(date), by = id]
df <- merge(df, tempdt, by = c("date", "event", "id"), all.x = TRUE)
df[, tae := ifelse(shift(event)==1, date - shift(date), tae), by = id]
修改
更一般的解决方案
df <- data.table(date=as.Date(c("06/07/2000","15/09/2000","15/10/2000","03/01/2001","17/03/2001", "18/03/2001",
"06/08/2010","15/09/2010","15/10/2010","03/01/2011","17/03/2011","19/03/2011"),
"%d/%m/%Y"),
event=c(1,0,0,0,0,0,1,1,1,0,1,0),id = c(rep(1,6),rep(5,6)))
##for event = 1 observations
tempdt <- df[event==1,]
tempdt[,tae := date - shift(date), by = id]
df <- merge(df, tempdt, by = c("date", "event", "id"), all.x = TRUE)
##for event = 0 observations
for(d in df[event==0, date]){
# print(as.Date(d, origin = "1970-01-01"))
df[date == d & event == 0, tae := as.Date(d, origin = "1970-01-01") -
max(df[date<d & event==1,date]), by = id]
}
编辑2
现在，必须有更快的方法来做到这一点，但如果第一次观察是event = 0，这不会导致任何警告
df <- data.table(date=as.Date(c("06/07/2000","15/09/2000","15/10/2000","03/01/2001","17/03/2001","06/08/2010","15/09/2010","15/10/2010","03/01/2011","17/03/2011"),
"%d/%m/%Y"), event=c(0,0,1,0,1, 1,0,0,0,1),id = c(rep(1,5),rep(2,5)))
tempdt <- df[event==1,]
tempdt[,tae := date - shift(date), by = id]
df <- merge(df, tempdt, by = c("date", "event", "id"), all.x = TRUE)
for(i in unique(df[,id])){
# print(i)
for(d in df[date>df[id == i & event==1,min(date)] & event==0, date]){
# print(as.Date(d, origin = "1970-01-01"))
df[id == i & date == d & event == 0,
tae := as.Date(d, origin = "1970-01-01") - max(df[date<d &
event==1,date])]
}
}